Solve Inequality: Algebraic Proof of a<b<c<1/12

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Homework Help Overview

The discussion revolves around proving an inequality involving a sequence of fractions and their relationship to the value 1/12. The original poster presents a sequence defined by natural numbers and seeks assistance in demonstrating a specific inequality.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the validity of the original poster's approach and suggest clarifications regarding the sequence of fractions. There are inquiries about how to continue the proof and suggestions for establishing upper bounds.

Discussion Status

Some participants have provided guidance on refining the approach and correcting specific elements of the original poster's work. There is an ongoing exploration of relationships between the products of the sequences mentioned, but no consensus has been reached on a definitive method or solution.

Contextual Notes

Participants are working within the constraints of the problem as presented, including the requirement to show the inequality without providing complete solutions. The discussion reflects a mix of interpretations and suggestions for further exploration.

Lizu
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hi! i need help for this inequality
1. ##a\in\mathbb{N}*~and~ \frac{a}{a+1}<\frac{a+1}{a+2}<\frac{a+2}{a+3}##
show that : ##\frac{1}{2}*\frac{4}{5}*...*\frac{2005}{2006}*\frac{2008}{2009}<\frac{1}{12}##

Here i have stoped. Please tell me if is corect what i have done so far and how to continue , or another idea to solve
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Use your inequality to obtain an upper bound for P_1^3.
 
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Hello Lizu, :welcome:

Good start. Except the last factors: P1 last one is ##{2008\over 2009} ## etc.
Make it so that your P1 is the product you are after
You can still show ##\ \ P_1P_2 P_3 = {1\over 2011}\ \ ## and ##\ \ P_1<P_2<P_3 \ \ ##.

So if you can show ##P_1^3 < P_1P_2P_3## you are in business !

[edit] ah, PA was faster. Nice exercise !
 
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Thank you !
 

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